The rate of change of x^{sin x} with respect | vedclass

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(A) Let $u = x^{\sin x}$. Taking logarithm on both sides,we get $\log u = \sin x \log x$.Differentiating with respect to $x$,we have:$\frac{1}{u} \fra

Vedclass · Accepted Answer

$\frac{x^{\sin x}\left(\frac{\sin x}{x}+\cos x \cdot \log x\right)}{(\sin x)^x(x \cdot \cot x+\log \sin x)}$

Vedclass · Answer

(A) Let $u = x^{\sin x}$. Taking logarithm on both sides,we get $\log u = \sin x \log x$.Differentiating with respect to $x$,we have:$\frac{1}{u} \frac{du}{dx} = \cos x \log x + \sin x \cdot \frac{1}{x} = \cos x \log x + \frac{\sin x}{x}$.Thus,$\frac{du}{dx} = x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right)$.Let $v = (\sin x)^x$. Taking logarithm on both sides,we get $\log v = x \log \sin x$.Differentiating with respect to $x$,we have:$\frac{1}{v} \frac{dv}{dx} = 1 \cdot \log \sin x + x \cdot \frac{1}{\sin x} \cdot \cos x = \log \sin x + x \cot x$.Thus,$\frac{dv}{dx} = (\sin x)^x (x \cot x + \log \sin x)$.The rate of change of $u$ with respect to $v$ is $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right)}{(\sin x)^x (x \cot x + \log \sin x)}$.

The rate of change of $x^{\sin x}$ with respect to $(\sin x)^{x}$ is

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